Lab2_n

MATH 152 Lab 2 Samuel Molero, Angus Ladd, Karthink Nuti, Arlin Birkby Instructions: Complete the lab assignment in your assigned groups. Unless stated otherwise, your answers should be obtained using Python code. Do not modify the cell above, as it contains all the packages you will need. It is highly recommended to not use any additional packages. NOTE: If you took MATH 151 last semester, notice that the import statement for SymPy is di ff erent- for each SymPy command you use, you have to preface it with "sp." For example, "symbols('x')" becomes "sp.symbols('x')". Except for plot and plot_parametric - you don't need to type "sp." for those. Question 1 1a The Total Volume of Ice Cream is: 2*pi 1b In   [19]: from sympy import * from sympy.plotting import ( plot , plot_parametric ) In   [24]: #start code here y = symbols ( 'y' , real = True ) r_squared = 1 - y ** 2 #display(r_squared) formula_1 = pi * r_squared Volume_top = integrate ( formula_1 ,( y , 0 , 1 )) #display(Volume_top) r_squared2 = ( y * Rational ( 1 , 4 )) ** Rational ( 2 ) #display(r_squared2) formula_2 = pi * r_squared2 Volume_bottom = integrate ( formula_2 ,( y , 0 , 4 )) #display(Volume_bottom) Total_Volume = Volume_top + Volume_bottom print ( 'The Total Volume of Ice Cream is:' , Total_Volume ) Lab2temp_152_23C http://localhost:8888/nbconvert/html/Desktop/152/Lab... 1 of 4 9/12/23, 10:04 PM

The radius must be: 1.62726396978388 in order for the volume to be 14.500 0000000000 Question 2 2a In   [25]: #start code here x = symbols ( 'x' , Real = True , positive = True ) vol_top = pi * Rational ( 2 , 3 ) * x vol_bot = pi * Rational ( 4 , 3 ) * x ** 2 #display(vol_top+vol_bot) equation_new = vol_top + vol_bot - 14.5 soln = solve ( equation_new , x ) #print(soln) volume_checkingwork = equation_new . subs ( x , soln [ 0 ]) check = volume_checkingwork + 14.5 #print(check.evalf()) print ( 'The radius must be:' , * soln , "in order for the volume to be" , " " , In   [9]: #start code here x = symbols ( 'x' , real = True ) f_x = 2 * x ** 2 g_x = abs (( x ** 2 ) - 9 ) plot_1 = plot (( f_x ,( x , - 5 , 5 )),( g_x ,( x , - 5 , 5 )), ylim = ( - 10 , 50 )) Lab2temp_152_23C http://localhost:8888/nbconvert/html/Desktop/152/Lab... 2 of 4 9/12/23, 10:04 PM

the area under the curve is: this represents the net change in the population from 0 to 5 months Question 4 4a 4b 4c In   [15]: #start code here t = symbols ( 't' , real = True ) birth = (( 3.941109 ) * E ** ( 0.01199 * t )) death = ( 2.744248 ) * E ** ( 0.009 * t ) eq = ( birth - death ) Area = integrate ( eq ,( t , 0 , 5 )) print ( 'the area under the curve is: ' ) display ( Area . evalf ()) print ( "this represents the net change in the population from 0 to 5 months 6.27355000167739 In   [57]: # Write your answers as comments. Make sure all the text is visible at once # # Compared to the result obtained fin part (a) its a good estimate of how th # as month progresses, if for part a we were to integrate from intervals 0, # a result close to the actual estimate. # In   [28]: #start code here c = symbols ( 'C' , real = True ) x1 = tan ( x ) ** 2 display ( integrate ( x1 ) + c ) C − x + sin ( x ) cos ( x ) In   [29]: #start code here x2 = x * tan ( x ** 2 ) ** 2 display ( integrate ( x2 ) + c ) C − + x 2 2 tan ( x 2 ) 2 In   [11]: # Write your answers as comments. Make sure all the text is visible at once # for part (a) you can simplify using trigonimetry and then integrate # for part (b) you can implement the u sustitution method # Lab2temp_152_23C http://localhost:8888/nbconvert/html/Desktop/152/Lab... 4 of 4 9/12/23, 10:04 PM